Question

If $y=\sqrt{x}+\frac{1}{\sqrt{x}}$, then show that $2x\frac{dy}{dx}+y=2\sqrt{x}$

Answer 1

$y=\sqrt{x}+\frac{1}{\sqrt{x}}$

$2x\frac{dy}{dx}+y=2\sqrt{x}------------\left(i\right)$

therefore $\frac{dy}{dx}=\frac{d\sqrt{x}}{dx}+\frac{d}{dx\sqrt{x}}$

$=>\frac{1}{2\sqrt{x}}-\frac{1}{2x\sqrt{x}}------------\left(ii\right)$

Now put the value of $\frac{dy}{dx}$ in $\left(i\right)$ we get

$=>2x[\frac{1}{2\sqrt{x}}-\frac{1}{2x\sqrt{x}}]+\sqrt{x}+\frac{1}{\sqrt{x}}$

$=>\sqrt{x}-\frac{1}{\sqrt{x}}+\sqrt{x}+\frac{1}{\sqrt{x}}$

$=>2\sqrt{x}$